
where \({\rm i}=\sqrt{-1}\) , the second step holds by using the identity \(\det {\left( {{\bf {I}} - {\bf {X}}} \right)^{ - a}} = {}_1{F_0}\left( {a;{\bf {X}}} \right)\)  from [1]}, and \({}_p{F_q}\left( {{\bf {X}}} \right)\)  denotes the hypergeometric function of one Hermitian matrix [2]}. Hence, the CDF of \(X_k\) , i.e., \(F_{{{X_k}}} \left( x \right)\) , can be obtained by taking expectation of (REF ) over the distribution of \({{{\bf {\tilde{Z}}}}_w}\)  as
\(&F_{{{X_k}}} \left( x \right)= {{\mathbb {E}}_{{{{\bf {\tilde{Z}}}}_w}}}\left\lbrace  {{F_{\left. {{X_k}} \right|{{\bf {R}}_r}^{1/2}{\bf {Q}}{{\bf {R}}_r}^{1/2}}}\left( {\left. x \right|{{\bf {R}}_r}^{1/2}{\bf {Q}}{{\bf {R}}_r}^{1/2}} \right)} \right\rbrace \\&= \frac{1}{{2\pi {\rm i}}}\int \nolimits _{a - {\rm i}\infty }^{a + {\rm i}\infty } {\frac{{{e^{sx}}}}{s}{{\rm {E}}_{{{{\bf {\tilde{Z}}}}_w}}}\left\lbrace  {{}_1{F_0}\left( {1; - s{{\bf {R}}_r}^{1/2}{\bf {Q}}{{\bf {R}}_r}^{1/2}} \right)} \right\rbrace ds}\\&= \frac{1}{{2\pi {\rm i}}}\int \nolimits _{a - {\rm i}\infty }^{a + {\rm i}\infty } {\frac{{{e^{sx}}}}{s}{{\rm {E}}_{{{{\bf {\tilde{Z}}}}_w}}}\left\lbrace  {{}_1{F_0}\left( {1; - s{{\bf {R}}_r}{\bf {Q}}} \right)} \right\rbrace ds}.\) 
